Question: Is ${573039}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {573039}= &&{5}\cdot100000+ \\&&{7}\cdot10000+ \\&&{3}\cdot1000+ \\&&{0}\cdot100+ \\&&{3}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {573039}= &&{5}(99999+1)+ \\&&{7}(9999+1)+ \\&&{3}(999+1)+ \\&&{0}(99+1)+ \\&&{3}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {573039}= &&\gray{5\cdot99999}+ \\&&\gray{7\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {5}+{7}+{3}+{0}+{3}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${573039}$ is divisible by $3$ if ${ 5}+{7}+{3}+{0}+{3}+{9}$ is divisible by $3$ Add the digits of ${573039}$ $ {5}+{7}+{3}+{0}+{3}+{9} = {27} $ If ${27}$ is divisible by $3$ , then ${573039}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${573039}$ must also be divisible by $3$.